\(\int \frac {(a+b x^2) (c+d x^2)}{(e+f x^2)^4} \, dx\) [8]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=-\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \]

[Out]

-1/6*(-c*f+d*e)*x*(b*x^2+a)/e/f/(f*x^2+e)^3-1/24*(3*b*e*(c*f+d*e)-a*f*(5*c*f+d*e))*x/e^2/f^2/(f*x^2+e)^2+1/16*
(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*x/e^3/f^2/(f*x^2+e)+1/16*(b*e*(c*f+d*e)+a*f*(5*c*f+d*e))*arctan(x*f^(1/2)/e^(1
/2))/e^(7/2)/f^(5/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {540, 393, 205, 211} \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {\arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (a f (5 c f+d e)+b e (c f+d e))}{16 e^{7/2} f^{5/2}}+\frac {x (a f (5 c f+d e)+b e (c f+d e))}{16 e^3 f^2 \left (e+f x^2\right )}-\frac {x (3 b e (c f+d e)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (a+b x^2\right ) (d e-c f)}{6 e f \left (e+f x^2\right )^3} \]

[In]

Int[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

-1/6*((d*e - c*f)*x*(a + b*x^2))/(e*f*(e + f*x^2)^3) - ((3*b*e*(d*e + c*f) - a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2
*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f
*(d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 540

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*n*(p + 1))), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x
^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))
*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {\int \frac {-a (d e+5 c f)-3 b (d e+c f) x^2}{\left (e+f x^2\right )^3} \, dx}{6 e f} \\ & = -\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{\left (e+f x^2\right )^2} \, dx}{8 e^2 f^2} \\ & = -\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \int \frac {1}{e+f x^2} \, dx}{16 e^3 f^2} \\ & = -\frac {(d e-c f) x \left (a+b x^2\right )}{6 e f \left (e+f x^2\right )^3}-\frac {(3 b e (d e+c f)-a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {(b e-a f) (d e-c f) x}{6 e f^2 \left (e+f x^2\right )^3}+\frac {(b e (-7 d e+c f)+a f (d e+5 c f)) x}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {(b e (d e+c f)+a f (d e+5 c f)) x}{16 e^3 f^2 \left (e+f x^2\right )}+\frac {(b e (d e+c f)+a f (d e+5 c f)) \arctan \left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{5/2}} \]

[In]

Integrate[((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x]

[Out]

((b*e - a*f)*(d*e - c*f)*x)/(6*e*f^2*(e + f*x^2)^3) + ((b*e*(-7*d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(24*e^2*f^2
*(e + f*x^2)^2) + ((b*e*(d*e + c*f) + a*f*(d*e + 5*c*f))*x)/(16*e^3*f^2*(e + f*x^2)) + ((b*e*(d*e + c*f) + a*f
*(d*e + 5*c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(5/2))

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 e^{3} f^{2} \sqrt {e f}}\)	\(163\)
risch	\(\frac {\frac {\left (5 a c \,f^{2}+a d e f +b c e f +b d \,e^{2}\right ) x^{5}}{16 e^{3}}+\frac {\left (5 a c \,f^{2}+a d e f +b c e f -b d \,e^{2}\right ) x^{3}}{6 e^{2} f}+\frac {\left (11 a c \,f^{2}-a d e f -b c e f -b d \,e^{2}\right ) x}{16 e \,f^{2}}}{\left (f \,x^{2}+e \right )^{3}}-\frac {5 \ln \left (f x +\sqrt {-e f}\right ) a c}{32 \sqrt {-e f}\, e^{3}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) a d}{32 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b c}{32 \sqrt {-e f}\, f \,e^{2}}-\frac {\ln \left (f x +\sqrt {-e f}\right ) b d}{32 \sqrt {-e f}\, f^{2} e}+\frac {5 \ln \left (-f x +\sqrt {-e f}\right ) a c}{32 \sqrt {-e f}\, e^{3}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) a d}{32 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b c}{32 \sqrt {-e f}\, f \,e^{2}}+\frac {\ln \left (-f x +\sqrt {-e f}\right ) b d}{32 \sqrt {-e f}\, f^{2} e}\)	\(331\)

[In]

int((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x,method=_RETURNVERBOSE)

[Out]

(1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3*x^5+1/6*(5*a*c*f^2+a*d*e*f+b*c*e*f-b*d*e^2)/e^2/f*x^3+1/16*(11*a
*c*f^2-a*d*e*f-b*c*e*f-b*d*e^2)/e/f^2*x)/(f*x^2+e)^3+1/16*(5*a*c*f^2+a*d*e*f+b*c*e*f+b*d*e^2)/e^3/f^2/(e*f)^(1
/2)*arctan(f*x/(e*f)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (155) = 310\).

Time = 0.29 (sec) , antiderivative size = 642, normalized size of antiderivative = 3.75 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\left [\frac {6 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 16 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} - 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) - 6 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{96 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}, \frac {3 \, {\left (b d e^{3} f^{3} + 5 \, a c e f^{5} + {\left (b c + a d\right )} e^{2} f^{4}\right )} x^{5} - 8 \, {\left (b d e^{4} f^{2} - 5 \, a c e^{2} f^{4} - {\left (b c + a d\right )} e^{3} f^{3}\right )} x^{3} + 3 \, {\left (b d e^{5} + 5 \, a c e^{3} f^{2} + {\left (b d e^{2} f^{3} + 5 \, a c f^{5} + {\left (b c + a d\right )} e f^{4}\right )} x^{6} + {\left (b c + a d\right )} e^{4} f + 3 \, {\left (b d e^{3} f^{2} + 5 \, a c e f^{4} + {\left (b c + a d\right )} e^{2} f^{3}\right )} x^{4} + 3 \, {\left (b d e^{4} f + 5 \, a c e^{2} f^{3} + {\left (b c + a d\right )} e^{3} f^{2}\right )} x^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) - 3 \, {\left (b d e^{5} f - 11 \, a c e^{3} f^{3} + {\left (b c + a d\right )} e^{4} f^{2}\right )} x}{48 \, {\left (e^{4} f^{6} x^{6} + 3 \, e^{5} f^{5} x^{4} + 3 \, e^{6} f^{4} x^{2} + e^{7} f^{3}\right )}}\right ] \]

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="fricas")

[Out]

[1/96*(6*(b*d*e^3*f^3 + 5*a*c*e*f^5 + (b*c + a*d)*e^2*f^4)*x^5 - 16*(b*d*e^4*f^2 - 5*a*c*e^2*f^4 - (b*c + a*d)
*e^3*f^3)*x^3 - 3*(b*d*e^5 + 5*a*c*e^3*f^2 + (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e
^4*f + 3*(b*d*e^3*f^2 + 5*a*c*e*f^4 + (b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4*f + 5*a*c*e^2*f^3 + (b*c + a*d)*e^
3*f^2)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x^2 + e)) - 6*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c +
a*d)*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e^5*f^5*x^4 + 3*e^6*f^4*x^2 + e^7*f^3), 1/48*(3*(b*d*e^3*f^3 + 5*a*c*e*f^5 +
(b*c + a*d)*e^2*f^4)*x^5 - 8*(b*d*e^4*f^2 - 5*a*c*e^2*f^4 - (b*c + a*d)*e^3*f^3)*x^3 + 3*(b*d*e^5 + 5*a*c*e^3
*f^2 + (b*d*e^2*f^3 + 5*a*c*f^5 + (b*c + a*d)*e*f^4)*x^6 + (b*c + a*d)*e^4*f + 3*(b*d*e^3*f^2 + 5*a*c*e*f^4 +
(b*c + a*d)*e^2*f^3)*x^4 + 3*(b*d*e^4*f + 5*a*c*e^2*f^3 + (b*c + a*d)*e^3*f^2)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)
*x/e) - 3*(b*d*e^5*f - 11*a*c*e^3*f^3 + (b*c + a*d)*e^4*f^2)*x)/(e^4*f^6*x^6 + 3*e^5*f^5*x^4 + 3*e^6*f^4*x^2 +
e^7*f^3)]

Sympy [A] (verification not implemented)

Time = 2.98 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.83 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (- e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{e^{7} f^{5}}} \cdot \left (5 a c f^{2} + a d e f + b c e f + b d e^{2}\right ) \log {\left (e^{4} f^{2} \sqrt {- \frac {1}{e^{7} f^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a c f^{4} + 3 a d e f^{3} + 3 b c e f^{3} + 3 b d e^{2} f^{2}\right ) + x^{3} \cdot \left (40 a c e f^{3} + 8 a d e^{2} f^{2} + 8 b c e^{2} f^{2} - 8 b d e^{3} f\right ) + x \left (33 a c e^{2} f^{2} - 3 a d e^{3} f - 3 b c e^{3} f - 3 b d e^{4}\right )}{48 e^{6} f^{2} + 144 e^{5} f^{3} x^{2} + 144 e^{4} f^{4} x^{4} + 48 e^{3} f^{5} x^{6}} \]

[In]

integrate((b*x**2+a)*(d*x**2+c)/(f*x**2+e)**4,x)

[Out]

-sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(-e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ sqrt(-1/(e**7*f**5))*(5*a*c*f**2 + a*d*e*f + b*c*e*f + b*d*e**2)*log(e**4*f**2*sqrt(-1/(e**7*f**5)) + x)/32
+ (x**5*(15*a*c*f**4 + 3*a*d*e*f**3 + 3*b*c*e*f**3 + 3*b*d*e**2*f**2) + x**3*(40*a*c*e*f**3 + 8*a*d*e**2*f**2
+ 8*b*c*e**2*f**2 - 8*b*d*e**3*f) + x*(33*a*c*e**2*f**2 - 3*a*d*e**3*f - 3*b*c*e**3*f - 3*b*d*e**4))/(48*e**6*
f**2 + 144*e**5*f**3*x**2 + 144*e**4*f**4*x**4 + 48*e**3*f**5*x**6)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {{\left (b d e^{2} + b c e f + a d e f + 5 \, a c f^{2}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{2}} + \frac {3 \, b d e^{2} f^{2} x^{5} + 3 \, b c e f^{3} x^{5} + 3 \, a d e f^{3} x^{5} + 15 \, a c f^{4} x^{5} - 8 \, b d e^{3} f x^{3} + 8 \, b c e^{2} f^{2} x^{3} + 8 \, a d e^{2} f^{2} x^{3} + 40 \, a c e f^{3} x^{3} - 3 \, b d e^{4} x - 3 \, b c e^{3} f x - 3 \, a d e^{3} f x + 33 \, a c e^{2} f^{2} x}{48 \, {\left (f x^{2} + e\right )}^{3} e^{3} f^{2}} \]

[In]

integrate((b*x^2+a)*(d*x^2+c)/(f*x^2+e)^4,x, algorithm="giac")

[Out]

1/16*(b*d*e^2 + b*c*e*f + a*d*e*f + 5*a*c*f^2)*arctan(f*x/sqrt(e*f))/(sqrt(e*f)*e^3*f^2) + 1/48*(3*b*d*e^2*f^2
*x^5 + 3*b*c*e*f^3*x^5 + 3*a*d*e*f^3*x^5 + 15*a*c*f^4*x^5 - 8*b*d*e^3*f*x^3 + 8*b*c*e^2*f^2*x^3 + 8*a*d*e^2*f^
2*x^3 + 40*a*c*e*f^3*x^3 - 3*b*d*e^4*x - 3*b*c*e^3*f*x - 3*a*d*e^3*f*x + 33*a*c*e^2*f^2*x)/((f*x^2 + e)^3*e^3*
f^2)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )}{\left (e+f x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^3}-\frac {x\,\left (b\,d\,e^2-11\,a\,c\,f^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e\,f^2}+\frac {x^3\,\left (5\,a\,c\,f^2-b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{6\,e^2\,f}}{e^3+3\,e^2\,f\,x^2+3\,e\,f^2\,x^4+f^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (5\,a\,c\,f^2+b\,d\,e^2+a\,d\,e\,f+b\,c\,e\,f\right )}{16\,e^{7/2}\,f^{5/2}} \]

[In]

int(((a + b*x^2)*(c + d*x^2))/(e + f*x^2)^4,x)

[Out]

((x^5*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/(16*e^3) - (x*(b*d*e^2 - 11*a*c*f^2 + a*d*e*f + b*c*e*f))/(16
*e*f^2) + (x^3*(5*a*c*f^2 - b*d*e^2 + a*d*e*f + b*c*e*f))/(6*e^2*f))/(e^3 + f^3*x^6 + 3*e^2*f*x^2 + 3*e*f^2*x^
4) + (atan((f^(1/2)*x)/e^(1/2))*(5*a*c*f^2 + b*d*e^2 + a*d*e*f + b*c*e*f))/(16*e^(7/2)*f^(5/2))